At what temperature will the pressure of a gas at `0^(@)C` doubles itself when volume remains constant?

Explanation

Given `: P_(1) = P` atm , `T_(1) = 273 K , P_(2) = 2P` atm , `T_(2)` = ?
Using Gay Lussac.s law `(P_(1))/(T_(1)) = (P_(2))/(T_(2)) implies T_(2) = (P_(2) T_(1))/(P_(1)) = (2 P "atm" xx 273 K)/(P "atm") = 546 K (273 ^(@) C)`